Large Deviations Problem

Question

Let $\left(X_n\right)_{n\geq 1}$ be i.i.d random variables on $\left(\Omega,\mathcal A, \mathbb P\right)$, $X_1$ with mean $\mu$, and

$$ L(\lambda) = \begin{cases} \log\mathbb E\left(e^{\lambda X_1}\right)<\infty, & \text{if }\mathbb E\left(e^{\lambda X_1}\right)<\infty \\ +\infty, & \text{otherwise, } \end{cases} $$

Show that for any $\lambda \in \mathbb R$, $$L(\lambda)\geq \lambda\mu$$

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The infinity case is clear, so we are trying to show that

$$\log\mathbb E\left(e^{\lambda X_1}\right)\geq \lambda \mu$$

so,

$$\mathbb E\left(e^{\lambda X_1}\right)\geq e^{\lambda\mu}$$

Jenson's Inequality states that

$$\mathbb E\left[\varphi(X_1)\right]\geq\varphi\left(\mathbb E\left[X_1\right]\right)$$

Answer 1

Jensen's inequality states that $\varphi(\mathbb E(Y))\leqslant \mathbb E(\varphi(Y))$ where $\varphi\colon\mathbb R_+\to\mathbb R_+$ is convex and $Y$ is a non-negative random variable.

Here, when $\mathbb E(e^{\lambda X_1})$ is finite, use Jensen's inequality with $\varphi(s):=\exp(s)$.